0
$\begingroup$

How to determine the radius of convergence of $\sum\frac{1}{n^{\sqrt{n}}}x^n$ without using l'hopital's rule?

I first tried root test, so $|a_n|^{\frac{1}{n}}=\frac{1}{n^{\frac{1}{\sqrt{n}}}}$ which I cannot see what the limit it has.

Then I tried ratio test, so $R=\lim|\frac{a_n}{a_{n+1}}|=\lim|\frac{{(n+1)}^{\sqrt{n+1}}}{n^{\sqrt{n}}}|$ in which I have no idea how to cancel term and determine the limit.

Does anyone have idea?

$\endgroup$
  • 1
    $\begingroup$ Where would L'Hopital be used here? $\endgroup$ – user228113 Jun 4 '17 at 16:54
  • $\begingroup$ What is $x_n$? You probably mean $x^n$. $\endgroup$ – Mark Viola Jun 4 '17 at 17:02
2
$\begingroup$

Hint:

$$n^{1/\sqrt n}=\left(n^{1/2\sqrt n}\right)^2=\left[\left(\sqrt n\right)^{1/\sqrt n}\right]^2\xrightarrow[n\to\infty]{}\ldots$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.